We first compute an explicit basis of the vector space
\(\operatorname{\mathsf {Sol}}\bigl ((n, d),v, x\bigr )\) .
Let
\(F(w) = \sum \limits _{(k, l) \in I} f_{(k, l)}(w) Z_{(k, l)}.\) 
The condition \(F \in \operatorname{\mathsf {Sol}}\bigl ((n, d), v, x\bigr )\)  yields the following constraints on
the coefficients \(f_{(k, l)}\) :
\(\left\lbrace \begin{array}{lcl}f_{(k, l)}(w+1) & = & \varepsilon ^k f_{(k, l)}(w) \\f_{(k, l)}(w+\tau ) & = & \varepsilon ^l \varphi _{x-v}(w) f_{(k, l)}(w).\end{array}\right.\) 
It follows from Lemma REF  that the vector space of holomorphic solutions of the system (REF )
is one-dimensional and generated by the function
\(f_{(k, l)}(w) = \exp \Bigl (- \frac{2 \pi i d}{n} kw\Bigr ) \theta \Bigl (w + \frac{1+ \tau }{2} + v - x - \frac{d}{n}(k \tau - l)\Bigr ).\) 
From Proposition REF  and formula (REF ) it follows that \(r_{((n, d), \tau )}(v; x, y)\)  is given by the following expression:
\(r_{((n, d), \tau )}(v; x, y) = \sum \limits _{(k, l) \in I}r_{(k, l)}(v; z) \,Z_{(k, l)}^\vee \otimes Z_{(k, l)},\) 
where \(z = y - x\)  and
\(r_{(k, l)}(v; z) = \exp \Bigl (-\frac{2 \pi i d}{n} k z\Bigr )\frac{\displaystyle \theta ^{\prime }\Bigl (\frac{1 + \tau }{2}\Bigr ) \theta \Bigl (z + v + \frac{1 + \tau }{2}- \frac{d}{n}(k\tau - l)\Bigr )}{\displaystyle \theta \Bigl (v + \frac{1 + \tau }{2}- \frac{d}{n}(k\tau - l)\Bigr ) \theta \Bigl (z + \frac{1 + \tau }{2}\Bigr )}.\) 
Relation (REF ) implies that
\(\frac{\displaystyle \theta ^{\prime }\Bigl (\frac{1 + \tau }{2}\Bigr ) \theta \Bigl (z + v + \frac{1 + \tau }{2}- \frac{d}{n}(k\tau - l)\Bigr )}{\displaystyle \theta \Bigl (v + \frac{1 + \tau }{2}- \frac{d}{n}(k\tau - l)\Bigr ) \theta \Bigl (z + \frac{1 + \tau }{2}\Bigr )} =\frac{\displaystyle \theta ^{\prime }\Bigl (\frac{1 + \tau }{2}\Bigr ) \bar{\theta }\Bigl (z+ v - \frac{d}{n}(k\tau - l)\Bigr )}{\displaystyle i \exp \Bigl (-\pi i \frac{\tau }{4}\Bigr )\bar{\theta }\Bigl (v -\frac{d}{n}(k\tau - l)\Bigr ) \bar{\theta }(z)}\) 
Moreover, it follows from (REF ) that \(\theta ^{\prime }\Bigl (\dfrac{1 + \tau }{2}\Bigr ) =i \exp \Bigl (-\pi i \dfrac{\tau }{4}\Bigr ) \bar{\theta }^{\prime }(0)\) . Hence, we get:
\(r_{(k, l)}(v; z) = \exp \Bigl (-\frac{2 \pi i d}{n} k z\Bigr )\frac{\displaystyle \bar{\theta }^{\prime }(0) \bar{\theta }\Bigl (z+ v - \frac{d}{n}(k\tau - l)\Bigr )}{\displaystyle \bar{\theta }\Bigl (v -\frac{d}{n}(k\tau - l)\Bigr ) \bar{\theta }(z)} \\=\exp \Bigl (-\frac{2 \pi i d}{n} k z\Bigr ) \sigma \Bigl (v -\frac{d}{n}(k\tau - l), z\Bigr ).\) 
Here we use the fact that
the Kronecker elliptic function \(\sigma (u, z)\)  defined by (REF ) satisfies the formula: \(\sigma (u, z)= \dfrac{\displaystyle \bar{\theta }^{\prime }(0) \bar{\theta }_1(u+z)}{\displaystyle \bar{\theta }(u) \bar{\theta }(z)}\) 
(see for instance [1]}).
We have a bijection \(\left\lbrace 1, \dots , n\right\rbrace  \longrightarrow \left\lbrace 0, \dots , n-1\right\rbrace , k \mapsto (n-k)\) .
Using this substitution as well as the identity
\(\sigma (u - d\tau , x) = \exp (2\pi i d z) \sigma (u, z),\) 
we end up with the expression (REF ), as asserted.
